Limiting reactant... ?

Question

One of the steps in them commericial process for converting ammonia to nitric acid involves the conversion of NH3 to NO

4NH3 + 5O2 =====>4NO + 6H2O

In a certain experiment 2.50 g of NH3 reacts with 2.85 g of O2

a Which reactant is the limiting reactant?

b) How many grams of NO form?

c)How much of the excess reactant remains after the limiting reactant is completely consumed?

I appreciate the help

10 points

Answer 1

idk

Answer 2

a) the limiting reactant is oxygen because :

moles: NH3 : O2

           0.147:0.0178 and since the molar ratio in the eq. is higher for O2 and it is less in the exp. then it limits the rxn.

b) molar ratio: NB choose O2 bcz it's the limiting reactant

     O2:NO

     5  : 4

0.0178: ?

cross multiply to find mole of NO then find mass of it using the formula:

mass=mole*molar mass

c) Calculate the mole of NH3 that reacts with the given moles of O2 then subtrat the ans from the given mole of NH3        

Answer 3

Moles NH3 = 2.50 g / 17.0307 g/mol =  0.147

Moles O2 = 2.85 g / 31.9988 g/mol = 0.0891

the ratio between NH3 and O2 is 4 : 5 and O2 is the limiting reactant

( 5 x 0.147 / 4 =  0.184 moles O2 needed)

the ratio between O2 ( limiting reactant ) and NO is 5 : 4

Moles NO = 0.0891 x 4 / 5 = 0.0713

Mass NO = 0.0713 mol x 30.0061 g/mol = 2.14 g

0.0891 x 4 / 5 = 0.0713 moles NH3 needed

moles NH3 in excess = 0.147 - 0.0713 = 0.0757

Mass Nh3 in excess = 0.0757 mol x 17.0307 g/mol =1.29 g